Question #346621

An aeroplane flies at a ground velocity (i.e. velocity relative to the ground) of 300 km/h N 30o W, in a wind blowing at a velocityof 50 km/h N 20o E. What is the velocity (speed and direction)of the plane relative to the ground? (Use a calculator and round the speed)



to the nearest km/h, and the corresponding angle to the nearest degree.)

1
Expert's answer
2022-06-01T10:30:32-0400
vp=300cos(90°+30°)i+300sin(90°+30°)j\vec{v_p}=300\cos(90\degree+30\degree)\vec{i}+300\sin(90\degree+30\degree)\vec{j}

vw=50cos(90°20°)i+50sin(90°20°)j\vec{v_w}=50\cos(90\degree-20\degree)\vec{i}+50\sin(90\degree-20\degree)\vec{j}

v=vp+vw=300cos(120°)i+300sin(120°)j\vec{v}=\vec{v_p}+\vec{v_w}=300\cos(120\degree)\vec{i}+300\sin(120\degree)\vec{j}


+50cos(70°)i+50sin(70°)j+50\cos(70\degree)\vec{i}+50\sin(70\degree)\vec{j}

v2=(300cos(120°)+50cos(70°))2|\vec{v}|^2=(300\cos(120\degree)+50\cos(70\degree))^2

+(300sin(120°)+50sin(70°))2+(300\sin(120\degree)+50\sin(70\degree))^2

111783.6283\approx111783.6283

v=v2334 km/h|\vec{v}|=\sqrt{|\vec{v}|^2}\approx334\ km/h


tanθ=300sin(120°)+50sin(70°)300cos(120°)+50cos(70°)\tan \theta=\dfrac{300\sin(120\degree)+50\sin(70\degree)}{300\cos(120\degree)+50\cos(70\degree)}

2.30846\approx-2.30846

θ=180°+tan1(2.30846)113°\theta=180\degree+\tan^{-1}(-2.30846)\approx113\degree

334 km/h334\ km/h, N 23°23\degreeW



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